Problem: Multiply the following rational expressions and simplify the result. $\dfrac{12y+8yz}{y^4+9y^3z+20y^2z^2} \cdot \dfrac{y^2-16z^2}{6z+9}=$
Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $12y+8yz$, of the first expression can be factored to $4y(3+2z)$ by factoring out $4y$. The denominator, $y^4+9y^3z+20y^2z^2$, of the first expression can be factored as $y^2(y+4z)(y+5z)$ by factoring out $y^2$ and using the sum-product pattern. The numerator, $y^2-16z^2$, of the second expression can be factored as $(y+4z)(y-4z)$ by the difference of squares pattern. The denominator, $6z+9$, of the second term can be factored as $3(2z+3)$ by factoring out $3$. Now the product looks as follows: $\dfrac{4y(3+2z)}{y^2(y+4z)(y+5z)} \cdot \dfrac{(y+4z)(y-4z)}{3(2z+3)}$ To multiply two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=} \dfrac{4y(3+2z)}{y^2(y+4z)(y+5z)} \cdot \dfrac{(y+4z)(y-4z)}{3(2z+3)}$ $\begin{aligned} &= \dfrac{4y(3+2z) \cdot (y+4z)(y-4z)}{y^2(y+4z)(y+5z) \cdot 3(2z+3)} &\text{Multiply across.}\\\\\\\\ &= \dfrac{4{\cancel{y}}{\cancel{(3+2z)}} {\cancel{(y+4z)}}(y-4z)}{{\cancel{y}} \cdot y{\cancel{(y+4z)}}(y+5z) 3{\cancel{(2z+3)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{4(y-4z)}{3y(y+5z)} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{4(y-4z)}{3y(y+5z)}$, which is equivalent to $\dfrac{4y-16z}{3y^2+15yz}$.